For measuring a figure in this form, it is plain we have this RULE. Multiply ONE OF THE LONGEST SIDES by the PERPENDICULAR A Ex. 1. Let A B or DC be 10 rods, and A E 8; required the area? 10 X 880 square rods, Ans. 2. Let the longer sides be 40 rods each, and perpendicular distance 39 rods; required the number of square rods? 3. How many acres in the above? Ans. 1560. Ans. 9A. 3R. 4. If the longer sides of a rhomboid be 50 rods each, and perpendicular 47, what is the area in acres, &c.? Ans. 14A. 2R. & 30sq. rods. 5. Given the base AB 8 rods, and perpendicular BC 6, to find the area. 8x648, and 482 gives 24 sq. rods, Ans. 6. Let the base be 18 rods, and perpen dicular 12; required the area. 7. Base 78, and perpendicular 45. 349. To measure an OBLIQUE-ANGLED TRIANGLE. B Ans. 2R. 28rds. Ans. 10A. 3R. 35rds. DEF. 1. A TRIANGLE is OBLIQUE, when not RIGHT-angled. 2. ANGLES are said to be OBTUSE when greater than right angles, and ACUTE when less. C D Before proceeding to apply the rule, we must evidently find the perpendicular CD; and in doing this, we must first find where, on the line AB, the perpendicular will fall. This may be done after the FOLLOWING MANNER: 349 a. As the WHOLE BASE is to the SUM OF THE TWO SHORTER SIDES : so is the DIFFERENCE of those sides to the DIFFERENCE OF THE *The longest side of a triangle is usually called the base, except in a right angled triangle, where the longest of the two legs is called the base. †The following is another method of finding the area of a triangle: From the HALF SUM of the three sides, take EACH SIDE severally, and note THE DIFFERENCES; then multiply the SAID HALF SUM and DIFFERENCES continually tegether, and the SQUARE ROOT of their product will be the AREA REQUIRED. It will be very good exercise for the learner to prove the above examples by this ule, SEGMENTS* of the base; then, by adding half this difference to half the base, we shall have the longer segment AD; and by subtracting said half difference from half the base, we have the shorter segment DB. After this is done, we may find the perpendicular by rule 2, Art 259. Applying the rule, 26 +22=48, and 26-224; then, as 40 : 48 :: 4:4.8, and of 40+ 2.4 (= 4.8 ÷ 2) = 22.4, the longer segment. Again, of 40- 2.4 17.6, the shorter segment. And by the rule, Art. 259, √222 17.6 13.2, the perpendicular. By the same rule, √262 — 22.42 = 13.2, as before. Consequently, 40 × 13.2÷2= 132, the area of the triangle. 9. What is the area of a triangle, the three sides being 13, 11, and 18? 20, 30, & 50? 26, 22, and 40? 33, 45, and 52 ? -63, 19, and 70 ? 32, 51, and 60? 72, 35, and 80 ? Ans. to the last, 1259.997984 rods. 350. To measure a TRAPEZIUM. DEF. A TRAPEZIUM is a figure of four UNEQUAL SIDES, and UNEQUAL angles. RULE. Dividing the trapezium as represented in the figure, we have two triangles, of which, the sum of the perpendiculars, multiplied into the DIAGONAL, gives the DOUBLE AREA. 10. Given the perpendiculars De & Bf=6 and 10, and diagonal AC 24, to find the A B areas. 6+10= -- 16, then 16 x 24= 384, and 3842=192, Ans. 11. If the perpendiculars be 19 and 27, and diagonal 45, what is the area Ans. 877.5. 12. Given the perpendiculars = 35 and 17, and diagonal 63, to find the area? Ans. 1638. 13. How many acres in the last, allowing the dimensions in rods? Ans. 10A. 4 38rds. 351. To measure a TRAPEZOID. DEF. A TRAPEZOID is the segment of a triangle, cut by a line parallel to the base. RULE. Multiply HALF THE SUM of the parallel sides by the PERPEN DICULAR DISTANCE. 14. Given the side AB 25, the side A CD 13, and perpendicular 8, to find D e B the area. 2513219, and 19 x 8=152, Ans. 15. If the parallel sides be 35 and 17, and perpendicular 20, what is the area? Ans. 250. 16. The two parallel sides being 79 and 50, and perpendicular 35, what is the area? Ans. 2257 A segment or frustum signifies a PIECE or PART of a thing. 352. To find the area of an IRREGULAR FIGURE, RULE. Divide the FIGURE into triangles, then measure EACH OF THE TRIANGLES, and the sum of their SEVERAL AREAS, will be the AREA REQUIRED. The figure ABCDEF being given, divide it into triangles by the diagonals AC, FC, and EC: then these triangles may be measured by letting fall perpendiculars, as Ad, Fc, Ca, Cb, and multipling these perpendiculars, by HALF THEIR RE SPECTIVE BASES. F E D 17. In the triangle ABC the base AB is 50, and perpendicular 19; in the triangle FCA the base FC is 52, and perpendicular 21; in the triangle ECF the the base EC is 55, and perpendicular 23; and in the triangle EDC the base ED is 59 and perpendicular Cb 20; required the area of the figure ABCDEF. 353. To measure a REGULAR POLYGON. Ans. 2243.5. DEF. A REGULAR POLYGON is a figure whose sides and an gles are ALL EQUAL. RULE. Multiply the length of ONE SIDE, by HALF THE NUMBER OF SIDES, and that product by the PER PENDICULAR. 18. Required the area of the five sided figure ABCDE, each side of which being 25 and perpendicular fg 17.2047737. 25 × 262; then 624 X 17.204773 107.298356, Ans. E D A g C B The names of several polygons, together with their superficial con tents, (the sides being ONE,) may be seen in the following Hence, if we square the side of a polygon, and multiply it by the MULTIPLIER, it will give the AREA. 354. To find the area of a CIRCLE.* RULE. MULTIPLY HALF THE DIAMETER INTO HALF THE CIRCUMFERENCE. 1. Diam. X 3.14159, or.31831, gives the circumference. 2. Circum. x .31831, or 3.141659, gives the diameter. 3. Diam. x .886227, or 1.128379, gives the side of an equal square. 4. Diam. X.866024, or.1547, the side of an inscribed equilateral triangle 5. Diam. X.70710678, or 6. Diam. x .785398, or 7. Circum. X.282094, or 1.414203, gives the side of an inscribed square. 1.273241, gives the area of the circle. 3.544907, gives the side of an equal square. 19. Required the area of a circle, whose diameter is 7 feet, and circumference 22? Ans. 384sq. ft. 355. As 7 is to 22, or more exactly, as 113 to 355, or by decimals, as 1 to 3.1416, so is the DIAMETER OF A CIRCLE to its CIRCUMFERENCE. 20. What is the circumference of a circle, whose diameter is 12? Ans. 37.71, 37.699, or 37.6992. 21. The diameter of a circle, whose circumference is 12? Ans. 3.8+. 22. What is the area of a circle, drawn within a square, containing 1 square rod? The diameter of the circle will be 1 and circumference 3.1416; therefore, Ans. 7854 nearly. 356. Hence, if the diameter of a circle be given, to find the AREA, RULE. MULTIPLY THE SQUARE OF THE DIAMETER BY .7854. 23. What is the area of a circle, whose diameter is 12? Ans. 113.0976. 24. whose diameter is 113 rods? Ans. 62A. 2R. 28.77sq. rods. NOTE 1. If the area be given to find the diameter, reverse the process. 2. The area of a half or semicircle is one half the area of the whole circle; and that of a quadrant is one fourth. 357. 25. Given the chord AH = 16 chains, and the chord AC = 10, to find the length of the arch line, ACH. RULE. From TWICE THE CHORD AC, take the CHORD AH, and of A the remainder added to TWICE AC, will give the ANSWER. 10 x 220, and 20-16=4; then of 42021 chains, Ans. 26. If, in the segment AHC, the chord AH be 216, do. of AC 126, what is the Jength of the arch line? Ans. 264. 358. 27. Given the versed sine DC 6 chains, and the chord AH 16, to find the diameter CE. Cir. Chord ence. Diame RULE. Divide the SQUARE of HALF THE CHORD by the VERSED SINE, and add the QUOTIENT to the SINE. 826 10.666 &c. which added to 6, gives 16.66666 &c. chains, Ans. 28. If the chord AB be 1869.5, and versed sine 423.5, what is the diame ter? Ans. 2486.6+. -359. Given the SECTOR of a circle, as AEBC (next page), to find the area. RULE. Multiply the RADIUS OF HALF DIAMETER, by HALF THE ARCH LINE. 29. Given radius AE= 72, chord AB 126, and AC 70, to find the area of the sector AEBC. Ans. 520776. 8. Circum. X .2756646, or 3.6275939, gives the side of an equilateral triangle inscribed. 9. Circum. x .225079, or÷4.442877, gives the side of an inscribed square. 12. Area X 12.56636217, or÷.079577525, gives the square of the circumference. 30. If radius AE be 84 chains, chord AB 16, and AC 10, required the area of the sector AEBC. Ans. 88.888888, &c. square chains. 31. If the chord Bg be 126, do. the chord BF 204, and radius EB or EA be 112, what is the area of the sector AEBF? 360. To find the area of a SEGMENT OF A CIRCLE, as ADBCA, or ADBFA. RULE. Find the area of the SECTOR, then add or subtract the area of the TRIANGLE formed by the CHORD and TWO RADII, as the case may require. 32. Required the area of the segment ADBC, whose chord AB is 172, do. AC or CB 104, and versed sine 58.49. NOTE. Here the area of the triangle must be subtracted from that of the sector. Ans. 7248.25. cir Ans. 28224 -cu Cherd B 33. If the chord AB, of the segment ADBFA, be 136 links, do. of BF 146, of Bg or gF 86, and radius 80, what is the area in acres, &c. NOTE. Here the area of the triangle must be added. Ans. 17A. IR. 9.48rds.. 361. To measure an ELLIPSIS or OVAL. DEF. AN ELLIPSIS or OVAL is a LONG ROUND, having two diameters; the longest of which is called the TRANSVERSE, and the shortest the CONJUGATE. RULE. The rectangle of the TWO DIAMETERS multiplied by .7854, will give the AREA required. 34. Required the area of an ellipsis, whose transverse and conjugate diameters are 88 and 72. 362. To measure the height of a TREE or OTHER INACCESSIBLE OBJECT. RULE. Set TWO STAKES at A and H, PARALLEL TO THE BODY OF THE TREE,. and ranging with the top C; then say, as AH: FD:: AB: GC, which added to BG, gives the height required. 35. If the shorter stake be 6 feet, distance between them 10, diff. of their lengths 8, and distance between the tallest and the tree be 35; required the height of the tree. E H Ans. 42 feet. 363. To find the breadth of a RIVER, or INACCESSIBLE MOAT. RULE. Begin at the BANK OF THE RIVER D, opposite some OBJECT (E), so that DA may be at RIGHT ANGLES with the bank of the river DC; and measure off any convenient distance to A; thence to B, making a right angle at A; thence (towards E) to the bank of the river at Ans. 4976.29+ C B |